What is the height of the plane when the decoy was released?

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The discussion centers on calculating the height of an airplane when a radar decoy is released while diving at a 30° angle. The plane's speed is converted to 70.83 m/s, and the horizontal distance to the impact point is 700 m. The time of flight is calculated to be 11.41 seconds, but the initial vertical velocity component must be considered for accurate height determination. The initial vertical velocity is derived from the plane's speed multiplied by the sine of the diving angle, which affects the final height calculation. The initial calculations are incorrect due to neglecting this vertical component, highlighting the importance of considering all velocity components in projectile motion.
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[SOLVED] projectile shot downward

A certain airplane has a speed of 255.0 km/h and is diving at an angle of 30.0° below the horizontal when a radar decoy is released (see Figure 4-36). The horizontal distance between the release point and the point where the decoy strikes the ground is 700 m. Neglect air resistance.

so Velocity in meters per second is 255*.2777=70.83m/s
Velocity in X direction is 70.83sin60=61.34m/s

Figure 4-36

(a) How high was the plane when the decoy was released?

t=m/v = 700/61.34 = 11.41s
Y=Yo+VoT+.5(a)(t^2)
Y=(0)+(0)(t)-4.9(11.41^2)
Y=638.1m which is wrong, can anyone help me out
 
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B-80 said:
t=m/v = 700/61.34 = 11.41s
Y=Yo+VoT+.5(a)(t^2)
Y=(0)+(0)(t)-4.9(11.41^2)
The y-component of the the initial velocity is not zero.
 
The initial vertical component of the velocity of the released item is not zero, due to the vis inertiae, since the aeroplane was traveling at two hundred and fifty-five kilometres per hour times the sine of thirty degrees vertically.
 
good point :p
 
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